4
$\begingroup$

I am learning about monoidal categories and I am a bit confused about the following: Suppose $(A,\otimes,I)$ is a monoidal category. What can be said about the opposite $A^{\text{op}}$? Is it immediately monoidal as well?

And if yes, what about extra structure that $A$ possibly has as a monoidal category, e.g. it is symmetric - is the opposite symmetric too?

$\endgroup$
1
  • 2
    $\begingroup$ Well, isomorphisms are still going to be isomorphisms in the opposite category, and $(A\times A)^{op}$ is the same thing as $A^{op}\times A^{op}$, so you still have a bifunctor. So all you need to check are the appropriate diagrams. $\endgroup$
    – Malice Vidrine
    Jun 24, 2016 at 8:29

1 Answer 1

Reset to default
11
$\begingroup$

Note that taking opposite category gives rise to a $2$-endofunctor: $$(-)^{op}:\mathrm{CAT}\rightarrow \mathrm{CAT}$$ on $2$-category of all categories contained in a given universe $\mathcal{U}$. This $2$-endofunctor is covariant on functors and contraviariant on natural transfromations.

Using this $2$-endofunctor you can argue as follows. If $$(\mathcal{C},\otimes,\alpha,\eta,\lambda,I)$$ is a monoidal category, then: $$(\mathcal{C}^{op},\otimes^{op},\alpha,\eta,\lambda,I)$$ satisfies axioms of monoidal category except for the fact that in Mac Lane's pentagon and identity triangles arrows are reversed. So if you are assuming that $\alpha$, $\eta$, $\lambda$ are natural isomorphisms(which is usually assumed c.f Categories for the working mathematician), then: $$(\mathcal{C}^{op},\otimes^{op},\alpha^{-1},\eta^{-1},\lambda^{-1},I)$$ is a monoidal category

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .